The logic of coherence

Abstract

To remedy the lack of precision attached to the concept of coherence, a plethora of probabilistic measures has been developed. To broaden the perspective, we do not focus on the differences between these quantitative but the differences between qualitative approaches to coherence by comparing three probabilistic definitions for the relation denoted by ‘coheres with’. To reveal the different logics underlying these relations, we introduce a considerable number of formal properties and examine whether the given coherence relations possess them. Among these properties are not only the classics reflexivity, symmetry and transitivity, but also a variety of features that concern propositions containing negation, conjunction and disjunction, as well as features that concern inconsistency, entailment and equivalence.

Appendix

(REF):

$P(x|x)$ = 1, and $P(x|{\bar{x}})$ = $P({\bar{x}}|x)$ = 0. Hence, the definientia of (IC), (AC) and therefore also (SC) are satisfied.

(SYM):

(AC), (IC) and thus also (SC) require bidirectional confirmation in random order.

(TRA):

This is a straightforward consequence of the fact that the underlying accounts of confirmation are not transitive (cf. Douven 2011; Roche and Shogenji 2014; Shogenji 2003).

(INC):

$P(x|x)$ = $P({\bar{x}}|{\bar{x}})$ = 1, and $P(x|{\bar{x}})$ = $P({\bar{x}}|x)$ = 0. Hence, since x and ${\bar{x}}$ are contingent and the probability function P is regular, $P(x|{\bar{x}})$ < P(x) < $P(x|x)$ and $P({\bar{x}}|x)$ < P(${\bar{x}}$) < $P({\bar{x}}|{\bar{x}})$. The definientia of (IC) and (AC), and thus also (SC), are therefore violated.

(INC′):

If x ⊥ y, then $P(x|y)$ = $P(y|x)$ = 0. Hence, $P(x|y)$ < P(x) and $P(x|y)$ < $P(x|\bar{y})$, as well as $P(y|x)$ < P(y) and $P(y|x)$ < $P(y|{\bar{x}})$. That is, the definientia of (IC) and (AC), and thus also (SC), are again violated.

(NSC):

First, if x ~_i y, then $P(x|y)$ > P(x), which is for all contingent propositions equivalent to $P(\bar{y}|{\bar{x}})$ > $P(\bar{y})$. Analogously, $P(y|x)$ > P(y) is equivalent to $P({\bar{x}}|\bar{y})$ > P(${\bar{x}}$). Thus, if x ~_i y, then also ${\bar{x}}$ ~_i $\bar{y}$, in order that ~_i satisfies (NSC). Secondly, probability distribution P₁ in Table 6 shows that neither ~_s nor ~_a satisfies (NSC). According to this distribution, $P_{1} (x|y)$ ≈ .731 > .637 ≈ P₁(x) > .5, and $P_{1} (y|x)$ ≈ .812 > .707 ≈ P₁(y) > .5. Hence, x ~_s y and therefore x ~_a y. However, $P_{1} (\bar{y}|{\bar{x}})$ ≈ .477 < .5 and hence neither ${\bar{x}}$ ~_a $\bar{y}$ nor ${\bar{x}}$ ~_s $\bar{y}$.

Table 6 Probability distributions for pairs of propositions

(CON):

If x ~_s y, then $P(y|x)$ > P(y) and $P(y|x)$ > $P(\bar{y}|x)$. Hence, $P(\bar{y}|x)$ < $P(\bar{y})$ and $P(\bar{y}|x)$ < $P(y|{\bar{x}})$}, and therefore x ≁_i $\bar{y}$ and x ≁_a $\bar{y}$ and thus also x ≁_s $\bar{y}$.

(CON′):

Consider probability distribution P₆ in Table 7. Here $P_{6}(x|y)$ = 1 > .5 = P₆(x) and $P_{6}(y|x)$ = .5 > .25 = P₆(y); hence, x ~_i y. Furthermore, $P_{6}(z|x)$ = .5 > .25 = P₆(z) and $P_{6}(x|z)$ = 1 > .5 = P₆(x), so that also x ~_i z. Nonetheless, y ⊥ z because P₆(y ∧ z) = 0, with the result that ~_i violates (CON′). On the other hand, if x ~_a y, then $P(y|x)$ > .5 and hence $P(\bar{y}|x)$ ≤ .5. Hence, since $P(\bar{y}|x)$ = P($\bar{y}$ ∧ $z|x$) + P($\bar{y}$ ∧ $\bar{z}|x$), P($\bar{y}$ ∧ $z|x$) ≤ .5. Finally, since y ⊥ z, z implies $\bar{y}$, in order that P($\bar{y}$ ∧ $z|x$) = $P(z|x)$ ≤ .5. That is, x ≁_a z and x ≁_s z, entailing that both relations satisfy (CON′).

Table 7 Probability distributions for triples of propositions, where a_i = 1 − Pi(x ∨ y ∨ z)

(WAC):

As to incremental coherence, since P is regular and x is contingent, P($x|x$ ∧ y) = 1 > P(x). Due to the symmetry of incremental confirmation, it also holds that P(x ∧ $y|x$) > P(x ∧ y); hence, x ~_i x ∧ y. As to absolute coherence, since P is regular and x is contingent, P($x|x$ ∧ y) = 1 > P(${\bar{x}}|x$ ∧ y). Moreover, P(x ∧ $y|x$) = $P(y|x)$ and P($\overline{x \wedge y} |x$) = P(${\bar{x}}$ ∨ $\bar{y}|x$) = $P(\bar{y}|x)$. Since x ~_a y, we also get $P(y|x)$ > $P(\bar{y}|x)$, so that P(x ∧ $y|x$) = $P(y|x)$ > $P(\bar{y}|x)$ = P($\overline{x \wedge y} |x$). Hence, x ~_a x ∧ y. Therefore, all three accounts satisfy (WAC).

(WOC):

As to incremental coherence, since P is regular and x is contingent, P(x ∨ $y|x$) = 1 > P(x). Due to the symmetry of incremental confirmation, we thus get x ~_i x ∨ y. As to absolute coherence, P(x ∨ $y|x$) = 1 > 0 = P($\overline{x \vee y} |x$). Furthermore, since x ~_a y, $P(x|y)$ > $P({\bar{x}}|y)$, and therefore P(x ∧ y) > P(${\bar{x}}$ ∧ y). Hence, P(x) = P(x ∧ y) + P(x ∧ $\bar{y}$) > P(${\bar{x}}$ ∧ y) + P(x ∧ $\bar{y}$) > P(${\bar{x}}$ ∧ y). Consequently, P($x|x$ ∨ y) = P(x)/P(x ∨ y) > P(${\bar{x}}$ ∧ y)/P(x ∨ y) = P(${\bar{x}}|x$ ∨ y). This means that x ~_a x ∨ y and x ~_s x ∨ y, so that all three accounts satisfy (WOC).

(WCC):

According to probability distribution P₂ in Table 6, P₂(x ∧ $y|x$) ≈ .516 > .5 > .188 ≈ P₂(x ∧ y), and P₂($x|x$ ∧ y) = 1 > .5 > P₂(x). Hence, x ~_i x ∧ y and x ~_a x ∧ y, so that also x ~_s x ∧ y. However, P₂($x|y$) ≈ .353 < 0.363 ≈ P₂(x) < .5. Therefore, neither x ~_i y nor x ~_a y, entailing that x ~_s y does not hold, too.

(WCD):

Consider again distribution P₂. P₂($x|x$ ∨ y) ≈ .514 > .5 > .363 ≈ P₂(x), and P₂(x ∨ $y|x$) = 1 > .707 ≈ P₂(x ∨ y) > .5. Thus, x ~_i x ∨ y and x ~_a x ∧ y, implying that also x ~_s x ∨ y But since P₂($x|y$) < P₂(x) < .5, x and y are assessed incoherent on all three accounts.

(EOC):

Since x is contingent and P is regular, P($x|x$ ∧ y) = 1 > P(x). Therefore, by the symmetry of incremental confirmation, P(x ∧ $y|x$) > P(x ∧ y), in order that ~_i satisfies (EOC). On the other hand, as is shown by distribution P₃ in Table 6, there are situations in which P₃(x ∧ $y|x$) = P₃(x ∧ $y|y$) ≈ .332 < P₃(x ∧ y) ≈ 0.147 < .5. Hence, neither x ~ x ∧ y nor y ~ x ∧ y for either ~_a or ~_s.

(EOD):

Since x ∨ y is contingent and P is regular, P(x ∨ $y|x$) = 1 > P(x ∨ y). Hence, by symmetry also P($x|x$ ∨ y) > P(x), entailing that x ~_i x ∨ y. But distribution P₄ in Table 6 proves possible that P($x|x$ ∨ y) = P($y|x$ ∨ y) = .5, so that neither x ~ x ∨ y nor y ~ x ∨ y for either ~_a or ~_s.

(CSC):

Consider probability distribution P₇ in Table 7. P₇($x|y$) ≈ .165 > .163 ≈ P₇(x), in order that by symmetry x ~_i y. However, P₇(x ∧ $z|y$ ∧ z) ≈ .003 < .006 ≈ P₇(x ∧ z), implying that x ∧ z ~_i y ∧ z does not hold. Furthermore, according to distribution P₈ in the same table, P₈($x|y$) ≈ .613 > .5 > .419 ≈ P₈(x) and P₈($y|x$) = 1 > .684 ≈ P₈(y) > .5; therefore x ~_s y. But P₈(x ∧ $z|y$ ∧ z) ≈ .333 < .5, with the result that both ~_a and ~_s violate (CSC).

(DWC):

To show that ~_a satisfies (DWC), assume that x ∨ z ≁_a y ∨ z because P(x ∨ $z|y$ ∨ z) = P(x ∧ y ∧ $z|y$ ∨ z) + P(x ∧ y ∧ $\bar{z}|y$ ∨ z) + P(x ∧ $\bar{y}$ ∧ $z|y$ ∨ z) + P(${\bar{x}}$ ∧ y ∧ $z|y$ ∨ z) + P(${\bar{x}}$ ∧ $\bar{y}$ ∧ $z|y$ ∨ z) ≤ .5. Since P(x ∧ $\bar{y}$ ∧ $\bar{z}|y$ ∨ z) = P(${\bar{x}}$ ∧ $\bar{y}$ ∧ $\bar{z}|y$ ∨ z) = 0, this implies that the remaining probability P(${\bar{x}}$ ∧ y ∧ $\bar{z}|y$ ∨ z) > .5. Hence, P(${\bar{x}}$ ∧ y ∧ $\bar{z}|y$) ≥ P(${\bar{x}}$ ∧ y ∧ $\bar{z}|y$ ∨ z) > .5. But this entails that P(x ∧ y ∧ $z|y$) + P(x ∧ y ∧ $\bar{z}|y$) = $P(x|y)$ ≤ .5, so that x ≁_a y. Since an analogous argument applies if x ∨ z ≁_a y ∨ z because P(y ∨ $z|x$ ∨ z) ≤ .5, ~_a satisfies (DWC). Furthermore, distribution P₇ in Table 7 proves that ~_i does not meet this constraint. Here, x ~_i y but P₇(x ∨ $z|y$ ∨ z) ≈ .730 < .732 ≈ P₇(x ∨ z), so that x ∨ z ≁_i y ∨ z. One can easily find similar distributions showing that ~_s also violates (DWC).

(CMC):

We have seen that (CSC) is violated by all coherence relations, that is, it is possible that x ~ y but x ∧ z ≁ y ∧ z. Since reflexivity implies that z ~ z, it is also possible that x ~ y and z ~ z but x ∧ z ≁ y ∧ z. Hence, if z′ = z, then it is possible that x ~ y and z ~ z′ while x ∧ z ~ y ∧ z′. Hence, (CMC) is also violated by all coherence relations.

(DMC):

By analogy with (CMC), the fact that ~_s and ~_i violate (DWC) entails that they also violate (DMC). To show that, although ~_a satisfied (DWC), it does not satisfy (DMC), let y be $\bar{y}$, z be y and z′ be ${\bar{x}}$, and assume that x ~_a $\bar{y}$ and y ~_a ${\bar{x}}$. Probability distribution P₉ in Table 7 provides an example because all relevant conditional probabilities for these pairs equal 1. However, it is clearly not the case that x ∨ y ~_a $\bar{y}$ ∨ ${\bar{x}}$.

(LAC):

Probability distribution P₁₀ in Table 7 shows that none of the coherence relations satisfies (LAC). P₁₀($x|z$) ≈ .586 > .555 ≈ P₁₀(x) > .5, P₁₀($z|x$) ≈ .581 > .551 ≈ P₁₀(z) > .5, P₁₀($y|z$) ≈ .619 > .587 ≈ P₁₀(y) > .5 and P₁₀($z|y$) ≈ .580 > .551 ≈ P₁₀(z) > .5. Hence, x ~_s z and y ~_s z. However, P₁₀(x ∧ $y|z$) ≈ .450 < .480 ≈ P₁₀(x ∧ $y|z$) < .5. Therefore, neither x ∧ y ~_i z nor x ∧ y ~_a z, and a fortiori not x ∧ y ~_s z.

(LOC):

An example showing that none of the accounts satisfies (LOC) is given by probability distribution P₁₁. P₁₁($x|z$) ≈ .535 > .532 ≈ P₁₁(x) > .5, P₁₁($z|x$) ≈ .501 > .5 > .498 ≈ P₁₁(z), P₁₁($y|z$) ≈ .531 > .527 ≈ P₁₁(y) > .5 and P₁₁($z|y$) ≈ .502 > .5 > .498 ≈ P₁₁(z). Accordingly, x ~_s z and y ~_s z. But P₁₁($z|x$ ∨ y) ≈ .493 < .498 ≈ P₁₁(z) < .5. Hence, it is not the case that x ∨ y ~ z for each of the three accounts.

(RAC):

Due to the symmetry of the coherence relations, (RAC) is equivalent to (LAC).

(ROC):

Due to the symmetry of the coherence relations, (ROC) is equivalent to (LOC).

(LLE):

This is a straightforward consequence of the fact that the laws governing probability are not affected when propositions are replaced by equivalent propositions.

(LRE):

Ditto.

(CEQ):

If x and y are equivalent, then $P(x|y)$ and $P(y|x)$ equal 1 and are thus smaller than .5. Moreover, since x and y are contingent and P is regular, it is also entailed that $P(x|y)$ > $P({\bar{x}}|y)$ and $P(y|x)$ > $P(\bar{y}|x)$. Hence, x ~ y for all three relations.

(CEN):

If x entails y, then $P(y|x)$ = 1. Hence, since y is contingent and P is regular, it is also implied that $P(y|x)$ > P(y). By symmetry we obtain $P(x|y)$ > P(x), with the result that x ~_i y. On the other hand, consider probability distribution P₅ in Table 6. Here x entails y because P₅($y|x$) = 1 with x and y being contingent. However, P₅($x|y$) = 1/9, so that neither x ~_a y nor x ~_s y.

(LMO):

According to probability distribution P₁₂ from Table 7, P₁₂($x|y$) ≈ .798 > .5 > .474 ≈ P₁₂(x) and P₁₂($y|x$) ≈ .604 > .5 > .359 ≈ P₁₂(y). Therefore, x ~ y for all three coherence relations. Furthermore, y ⊢ z because P₁₂($z|y$) = 1 with z and y being contingent. However, x ~ z holds for no relation because P₁₂($x|z$) ≈ .352 < .474 ≈ P₁₂(x) < .5.

(RMO):

Consider probability distribution P₁₃. P₁₃($x|y$) = 1 > .751 > .5 ≈ P₁₃(x) and P₁₃($y|x$) ≈ .670 > .503 ≈ P₁₃(y) > .5. Hence, x ~ y on all three accounts. Moreover, z ⊢ x because P₁₂($x|z$) = 1 with x and z being contingent. Nonetheless, we have P₁₃($y|z$) = 0 < .5 < .503 ≈ P₁₃(y), implying that y ~ z obtains for none of the accounts.